I'm 12 so please don't laugh at my math.

Let's assume that an object has KE expressed by velocity $v$ and mass $m$. If work $w$ has been done to this object such that it's new speed (same direction) is $s$. There are 2 ways(I know) to calculate $w$. The first is $ \frac {1}{2} ms^2 - \frac {1}{2} mv^2$ Now, I know what you are thinking, what other way do you need? Anyway, the other way is to(assuming you know distance and acceleration) $w=mda $ then $d= \frac {1}{2}at^2 + vt $

$\frac{2d}{a}= t^2 + \frac {2v}{a}t$

$\frac {2d}{a} +\frac {v^2}{a^2} = t^2 + \frac {2v}{a}t + \frac {v^2}{a^2}$

$\frac {2d}{a} +\frac {v^2}{a^2} = (t+\frac {v}{a})^2$

$\sqrt {\frac {2d}{a} +\frac {v^2}{a^2}} = t + \frac{v}{a}$

$\sqrt {\frac {2da+v^2}{a^2}}= t + \frac{v}{a}$

$\frac {\sqrt {2da+v^2}}{a} = t + \frac{v}{a}$

$\frac {\sqrt {2da+v^2}-v}{a} = t $

and the speed is a*t which is equal to

${\sqrt {2da+v^2}-v}$

remember that

$ da = \frac {w}{m}$

$ s = \sqrt {\frac {2w}{m}+v^2}-v$

$ (s+v)^2 = \frac{2w}{m} +v^2$

$ \frac{1}{2}m((s+v)^2-v^2) = w $

simplify and it will be

$ w= \frac {1}{2} ms^2 + \frac{1}{2} mvs$

which isn't quite the other formula. Why is it that my formula doesn't work?